Consider the function where the integral is along any rectifiable path from to . Clearly this is not a single-valued function of , as the value of the integral will depend on the winding number of the path around . Thus is defined up to an integral multiple of . Now consider the inverse function ; we have and therefore is a periodic function. (It's the exponential!)

Now consider the function . Now is defined up to where . Explain why, mimicking the construction of the inverse function as above, we do not obtain an elliptic function.

May 7th 2010, 12:59 PM

tonio

Quote:

Originally Posted by Bruno J.

Consider the function where the integral is along any rectifiable path from to . Clearly this is not a single-valued function of , as the value of the integral will depend on the winding number of the path around . Thus is defined up to an integral multiple of . Now consider the inverse function ; we have and therefore is a periodic function. (It's the exponential!)

Now consider the function . Now is defined up to where . Explain why, mimicking the construction of the inverse function as above, we do not obtain an elliptic function.

I may be off orbit by long miles, but let's give it a try:

Spoiler:

Indeed we get , so this function is doubly periodic (...really? Read on), and thus, what's lacking to consider it an elliptic function? Well, the periods must be a basis for the real dimensional linear space , and in this case we get , which screws the whole thing up.
The complete, long and deep explanation may be well beyond what I'd be willing to explain by this means: We need a free abelian group in which is also a maximal order there and etc.
Making it short, though, we can simply say: that the above ratio is real and not complex non-real means the function has actually just one single period (!) which, automatically, disqualifies it from being an elliptic function.

That's why I used instead of : so that the residue at (or times the residue, if you prefer) would be real.

So the two "periods" are indeed linearly independent over .

Sorry about the typo; corrected.

May 7th 2010, 01:13 PM

tonio

Quote:

Originally Posted by Bruno J.

Oh, I'm sorry, my post should have read

That's why I used instead of : so that the residue at (or times the residue, if you prefer) would be real.

So the two "periods" are indeed linearly independent over .

Sorry about the typo; corrected.

Spoiler:

Mind you, I did notice the thing and thought it is a little odd the second assumed period is pure imaginary, but of course I didn't check it (as I am not checking it, either(Giggle) ) .

Well, then the two periods are a basis for the complex, so then the only reason why that function wouldn't be an elliptic one I can think of right now is that the function isn't meromorphic on a fundamental parallelogram. Now, either the function is there holomorphic and thus bounded and thus constant (but STILL would be consider elliptic, imo), or else it has a singularity that it is not a pole....oh, I think I see now! Zero is there...hmmm.
Anyway, it's late here so I shall check this closer tomorrow, perhaps.

Tonio

May 10th 2010, 09:08 PM

Bruno J.

Here's a hint : the universal cover of the twice punctured plane is the disc...

May 12th 2010, 08:42 AM

Bruno J.

Here's the solution I know. I suspect there is a less technical one.

The function is well defined , i.e. it is well defined on the elliptic curve . Now suppose there were an analytic inverse function . Lifting to a map between the universal coverings (which are, respectively, for and the unit disc for the twice-punctured plane ), we obtain an analytic map , which, by Liouville's theorem, must be constant...